calcolate tramite un integrale il valore medio del quadrato della funzione seno (o coseno) su un periodo e poi fatene la radice quadrata

1. ### Break Down the Problem

To calculate the average value of the square of the sine function $\sin^2(x)$ (or cosine function $\cos^2(x)$) over one period, we will:

1. Identify the period of the function.
2. Set up the integral for averaging.
3. Calculate the integral.
4. Take the square root of the average value.

2. ### Relevant Concepts

The period of the sine and cosine functions is $2\pi$. The average value of a function $f(x)$ over the interval $[a, b]$ is given by the formula:

$\text{Average} = \frac{1}{b - a} \int_a^b f(x) \, dx$

For $f(x) = \sin^2(x)$ or $f(x) = \cos^2(x)$ over the interval $[0, 2\pi]$:

$\text{Average} = \frac{1}{2\pi} \int_0^{2\pi} \sin^2(x) \, dx$

3. ### Analysis and Detail

Step 1: Calculate the Integral

Using the identity:

$\sin^2(x) = \frac{1 - \cos(2x)}{2}$

we can rewrite the integral:

$\int_0^{2\pi} \sin^2(x) \, dx = \int_0^{2\pi} \frac{1 - \cos(2x)}{2} \, dx$

This can be split into two integrals:

$= \frac{1}{2} \int_0^{2\pi} 1 \, dx - \frac{1}{2} \int_0^{2\pi} \cos(2x) \, dx$

Calculating the first integral:

$\int_0^{2\pi} 1 \, dx = 2\pi$

Calculating the second integral (the integral of cosine over a full period is zero):

$\int_0^{2\pi} \cos(2x) \, dx = 0$

Putting it all together:

$\int_0^{2\pi} \sin^2(x) \, dx = \frac{1}{2} (2\pi) - \frac{1}{2} (0) = \pi$

Step 2: Average Value

Now substituting back to calculate the average:

$\text{Average} = \frac{1}{2\pi} \int_0^{2\pi} \sin^2(x) \, dx = \frac{1}{2\pi} \cdot \pi = \frac{1}{2}$

4. ### Verify and Summarize

The average value of $\sin^2(x)$ over one period $[0, 2\pi]$ is $\frac{1}{2}$. Taking the square root to find the final answer:

Final Answer

$\sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \approx 0.7071$

Thus, the average square root of the sine (or cosine) function over one period is $\frac{1}{\sqrt{2}}$.